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I'm looking for a proof of identity $\bf 4.223.3$ from Gradshteyn and Ryzhik's Tables of Integrals, Series and Products, namely $$ \int_{0}^{\infty}\ln\left(\,1 + 2{\rm e}^{-x}\cos\left(t\right) + {\rm e}^{-2x}\,\right)\,{\rm d}x = \frac{\pi^{2}}{6} - \frac{t^{2}}{2},\quad \left\vert\,t\,\right\vert < \pi $$ ( I suspect it holds for $\left\vert\, t\,\right\vert = \pi$ as well ).

  • Unlike the proceeding formulas $\bf 4.223.1$ and $\bf 4.223.2$ for $$ \int_{0}^{\infty}\ln\left(\,1 \pm {\rm e}^{-x}\,\right){\rm d}x, $$ its proof does not appear in part $9$ of Amdeberhan et al.'s Integrals in Gradshteyn and Ryzhik.
  • The reference in GR is for table $256$ of the $1867$ work Nouvelles tables d'intégrales définies ( large pdf ).
  • One reason I'm interested in the derivation is that the integral looks related to $$ \int_{0}^{\infty}\ln\left(\,1 - 2x{\rm e}^{-x} - {\rm e}^{-2x}\,\right)\,{\rm d}x, $$ which is equivalent to the integral asked in a previous unanswered question.
Felix Marin
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heiner
  • 1,178

5 Answers5

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Another approach is to notice that the integral is \begin{align*} & \int_0^{ + \infty } {\log \left| {1 + e^{ - (x + it)} } \right|^2 dx} = 2\int_0^{ + \infty } {\Re \log (1 + e^{ - (x + it)} )dx} \\ & = 2\Re \int_0^{ + \infty } {\log (1 + e^{ - (x + it)} )dx} = 2\Re \int_0^{ + \infty } {\sum\limits_{n = 1}^\infty {( - 1)^{n + 1} \frac{{e^{ - (x + it)n} }}{n}} dx} \\ & = 2\Re \sum\limits_{n = 1}^\infty {\frac{{( - 1)^{n + 1} e^{ - i nt } }}{n}\int_0^{ + \infty } {e^{ - nx} dx} } \\ & = 2\Re \sum\limits_{n = 1}^\infty {( - 1)^{n + 1} \frac{{e^{ - itn} }}{{n^2 }}} = 2\sum\limits_{n = 1}^\infty {( - 1)^{n + 1} \frac{{\cos (nt)}}{{n^2 }}} \end{align*} which is the Fourier series of $\frac{\pi^2}{6}-\frac{t^2}{2}$, $|t|<\pi$.

Gary
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7

Let the given integeral be $I(t)$. Differentiating wrt $t$,

$$I'(t) = \int_0^{\infty} \frac{-2e^{-x}\sin t}{1+2e^{-x}\cos t+e^{-2x}}dx$$

Substitute $e^{-x} = y \Rightarrow -e^{-x}dx = dy$, then

$$I'(t) = -\int_0^1 \frac{2\sin t}{1+2y\cos t +y^2}dy $$

Using $\sin^2 t + \cos^2 t = 1$,

$$I'(t) = -2\sin t\int_0^1 \frac{1}{(y + \cos t)^2 + \sin^2 t}dy = -2\sin t\left(\frac{1}{\sin t} \arctan\left(\frac{y + \cos t}{\sin t}\right)\right|_0^1$$

$$\Rightarrow I'(t) = -2\left(\arctan\left(\frac{1 + \cos t}{\sin t}\right) - \arctan\left(\frac{\cos t}{\sin t}\right)\right)$$

With $1 + \cos t = 2\cos^2 \frac{t}{2}$, $\sin t = 2\sin\frac{t}{2}\cos\frac{t}{2}$ and $\arctan t = \frac{\pi}{2} - \cot^{-1}t$, the above expression is

$$I'(t) = -t \Rightarrow I(t) = \frac{-t^2}{2} + c$$

To determine $c$, we need to calculate $I\left(\frac{\pi}{2}\right)$. From the original integral,

$$I\left(\frac{\pi}{2}\right) = \int_0^{\infty} \ln(1 + e^{-2x})dx = \frac{\pi^2}{24}$$

The above result is available in the linked pdf itself. Thereforce,

$$c = I\left(\frac{\pi}{2}\right) + \frac{1}{2}\left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{6}$$

Hence,

$$I(t) = \frac{\pi^2}{6} - \frac{t^2}{2}$$

Pranav Arora
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5

This is really a slight expansion on the comment by Masacroso: If you make the substitution $\exp(-t) = u,$ and then differentiate with respect to $t,$ you get a rational function integral, which is easy to evaluate. For the starting point, the obvious value is $t=\frac{\pi}{2},$ which gets rid of the cosine term.

Igor Rivin
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Note that

\begin{align} I(t)=&\int_0^\infty\ln(1 + 2e^{-x}\cos t + e^{-2x})\,dx \\ =&2Re \int_0^\infty\ln(1+e^{it-x})dx \overset{y=e^{it-x} } =2Re \int_{0}^{e^{it}} \frac{\ln(1+y)}y dy\\ I’(t) =& 2Re\>( i \ln (1+e^{it}))=2Re\> \left(i\ln\left(2\cos\frac t2 e^{i \frac t2}\right)\right)=-t \end{align} Then $$I(t)= I(0)+\int_0^t I’(s)ds = 2\int_0^1 \frac{\ln(1+y)}ydy-\int_0^t s\>ds= \frac{\pi^2}6 -\frac {t^2}2 $$

Quanto
  • 120,125
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Sketch for possible solution, too long for a comment: suppose we can differentiate under the integral sign, then if $I(t)$ is the original integral $$ \frac{d}{dt}I(t)=-2\sin t\int_0^{\infty }\frac1{e^x+2\cos t+e^{-x}}\mathop{}\!d x $$

Then setting $c_t:=2\cos t,\, s_t:=-2\sin t$ and with the change of variable $e^x=y$ you have that

$$ \frac{d}{dt}I(t)=s_t\int_1^{\infty }\frac1{y^2+c_t y+1}\mathop{}\!d y $$

Then the last integral is solvable in two different ways depending on the value of $c_t$. This would give a closed form for $\frac{d}{dt}I(t)$. Now the integral $I(0)$ seems attackable, thus you will have finally a first order linear differential equation, what "probably" can be solved explicitly.